-Y-TKMS 37 



suppose rlmt the tube in Fig. .9 starts from unit area on a eon- 

 ducting surface at 7. Then the charge at 7 is rr. Let the tuhe 

 be prolonged within the conductor and be made to end at s . 

 Then I, = 0. since there IN no field within the conductor. Then 

 !,>! = 4x<r. If >, is clo^e to the surface. bv the last proposition 

 it is parallel to the surface, and also has unit area, and therefore 

 Ij = 4 - 



Representation of intensity by the number of lines of 

 force through unit area perpendicular to the lines. Along 

 a tube of force intensity X cross-section is everywhere constant if 

 there i- no charge within the tu!x>. Draw a cross-section Sj where 

 the intensity is j p and imagine that through each unit area at S 1% l l 

 lines of tone pass along the tube. There will be in all IjSj lines 

 in the tul>e. Let S, be a section further on where the intensity is 

 I t . All the IjSj lines pass through S 2 , and the number through 



unit area of it will Ix* c"^^ '2' or W} ^ ^ equal to the 



1 ^ 



intensit \ - Hence if ue draw lines of force from any surface at 



the rate of I lines p< r unit area (the ana being perpendicular to the 



the mnnlxT passing through unit area at any point in 



their coins,- (the area being perjK-ndicular to the lines) will l>e equal 



e intensity at that area. 

 If the area of the surface from which we draw the lines is S. 



the total nuinlxT through it, f I</^. is termed the flux of force 



through S. The magnitude of the intensity, as well as its direc- 

 tion. is then tore indicated bv the lines of force when drawn on 

 this sea! ' re thevare closer together the field is stronger ; 



\\h--re they >}> n out the field is \uaker. 



Number of lines of force starting from unit quantity. 



If a charge q is at a point, the intensit \ OY< r the surface of a sphere 



vn with the point as centre is </ ;-. \Veinust then allow 



lines per unit area, or 4^;*-^ = 4-7r<y in all. That is, from 



ititv 7, \irti lines start, and from unit quantity 4"rr lines 

 start. 



If a tu!. ontodlM a charge 7 at any point the flux of 



, B we have seen, change> l>\ [-</. that is, the charge q adds 



its lines tn those already going along the tube. 



The normal component of the intensity N at any sur- 



face IB equal to the number of lines of force passing 



ough unit area of the surface. Let AB, Fig. J31, repre- 



d of the Mirf.ue. .nid let I, the intensity, make 



an a ith the normal to AH. Let AC represent the projec- 



AH perp. n.licular to I. Then AC - a COB 6. The number 



of I: \( I IcosOa. N.t. Hut the same 



number of lines pass through AC and AH. Hence the number 



through unit \H is N. 



